A sample of sodium hydroxide of volume 13.9 mL was titrated to the stoichiometric point with 20.6 mL of 0.209 M HCl(aq). What is the initial molarity of NaOH in the solution? Answer in units of M

Explanation:

NaOH   +   HCl   =   NaCl   +   H₂O

1 mole      1 mole

moles of hydrogen chloride reacted = .0206 x .209  

= 4.3054 x 10⁻³ moles .

moles of sodium hydroxide reacted will be same = 4.3054 x 10⁻³ moles

Let x  be the molarity of NaOH

13.9 mL of x molarity will contain

.0139 x moles

.0139 x = 4.3054 x 10⁻³

13.9 x 10⁻³ x =  4.3054 x 10⁻³

x = .3097 M .

= .31 M .

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explanation: